Stabilność – kryterium Nyquista

Kryterium Nyquista bada stabilność układu zamkniętego na podstawie przebiegu charakterystyki częstotliwościowej układu otwartego.

Z(s) Y(s)

-

Otwarcie sprzężenia zwrotnego

Y₁(s) E(s) -

Y₀(s)

$$G_{z}\left( s \right) = \frac{Y(s)}{Y_{0}(s)} = \frac{R(s) \bullet G_{0}(s)}{1 + R(s) \bullet G_{0}(s)}$$

O stabilności układu decyduje rozkład miejsc zerowych równania charakterystycznego. Wszystkie miejsca zerowe muszą leżeć lewej półpłaszczyźnie płaszczyzny zespolonej – części rzeczywiste miejsc zerowych muszą być mniejsze od zera. Stąd należy rozwiązać następujące równanie:

M_z(s) = 1 + R(s) • G₀(s) = 0

R(s) • G₀(s) = −1

Rozwiązaniem tego równania na płaszczyźnie zespolonej jest przebieg charakterystyki częstotliwościowej:

R(s) • G₀(s) = ( − 1,  j0)

Jeżeli charakterystyka częstotliwościowa układu otwartego nie obejmuje punktu (-1,j0), to układ po zamknięciu będzie stabilny.

Jeżeli charakterystyka częstotliwościowa układu otwartego obejmuje punkt (-1,j0) lub przechodzi przez niego, to układ po zamknięciu będzie niestabilny lub na granicy stabilności.

Badanie układu otwartego mówi nam o stabilności układu regulacji po zamknięciu pętli sprzężenia zwrotnego.

Przykład:

$$G\left( s \right) = \frac{k}{\left( T_{1}s + 1 \right)\left( T_{2}s + 1 \right)\left( T_{3}s + 1 \right)\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

R(s) = k_p

$$G_{\text{uo}}\left( s \right) = \frac{kk_{p}}{\left( T_{1}s + 1 \right)\left( T_{2}s + 1 \right)\left( T_{3}s + 1 \right)\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

$$G_{\text{uo}}\left( s \right) = \frac{kk_{p}}{\left\lbrack T_{1}T_{2}s^{2} + \left( T_{1} + T_{2} \right)s + 1 \right\rbrack\left( T_{3}s + 1 \right)\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

$$G_{\text{uo}}\left( s \right) = \frac{kk_{p}}{\left\lbrack T_{1}T_{2}T_{3}s^{3} + \left( T_{1}T_{2} + T_{1}T_{3} + T_{2}T_{3} \right)s^{2} + \left( T_{1} + T_{2} + T_{3} \right)s + 1 \right\rbrack\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

$$G_{\text{uo}}\left( s \right) = \frac{kk_{p}}{\begin{bmatrix} T_{1}T_{2}T_{3}T_{4}s^{4} + \left( T_{1}T_{2}T_{3} + T_{1}T_{2}T_{4} + {T_{1}T_{3}T}_{4} + T_{2}T_{3}T_{4} \right)s^{3} + \\ \left( T_{1}T_{2} + T_{1}T_{3} + T_{2}T_{3} + T_{1}T_{4} + T_{2}T_{4} + T_{3}T_{4} \right)s^{2} \\ + \left( T_{1} + T_{2} + T_{3} + T_{4} \right)s + 1 \\ \end{bmatrix}\left( T_{5}s + 1 \right)}$$

$$G_{\text{uo}}\left( s \right) = \frac{kk_{p}}{\begin{bmatrix} T_{1}T_{2}T_{3}T_{4}T_{5}s^{5} + \\ \left( T_{1}T_{2}T_{3}T_{4} + T_{1}T_{2}T_{3}T_{5} + T_{1}T_{2}T_{4}T_{5} + {T_{1}T_{3}T}_{4}T_{5} + T_{2}T_{3}T_{4}T_{5} \right)s^{4} + \\ \left( T_{1}T_{2}T_{5} + T_{1}T_{3}T_{5} + {T_{2}T_{3}T}_{5} + T_{1}T_{4}T_{5} + T_{2}T_{4}T_{5} + T_{3}T_{4}T_{5} + T_{1}T_{2}T_{3} + T_{1}T_{2}T_{4} + {T_{1}T_{3}T}_{4} + T_{2}T_{3}T_{4} \right)s^{3} \\ + \left( T_{1}T_{5} + T_{2}T_{5} + T_{3}T_{5} + T_{4}T_{5} + T_{1}T_{2} + T_{1}T_{3} + T_{2}T_{3} + T_{1}T_{4} + T_{2}T_{4} + T_{3}T_{4} \right)s^{2} \\ + \left( T_{1} + T_{2} + T_{3} + T_{4} + T_{5} \right)s + 1 \\ \end{bmatrix}}$$

Dane:

k = 1,   T₁ = 1,   T₂ = 2,   T₃ = 3,   T₄ = 4,   T₅ = 5

$$G_{\text{uo}}\left( s \right) = \frac{k_{p}}{\begin{bmatrix} 120s^{5} + \left( 24 + 30 + 40 + 60 + 120 \right)s^{4} + \\ \left( 10 + 15 + 30 + 20 + 40 + 60 + 6 + 8 + 12 + 24 \right)s^{3} \\ + \left( 5 + 10 + 15 + 20 + 2 + 3 + 6 + 4 + 8 + 12 \right)s^{2} \\ + \left( 1 + 2 + 3 + 4 + 5 \right)s + 1 \\ \end{bmatrix}}$$

$$G_{\text{uo}}\left( s \right) = \frac{k_{p}}{\begin{bmatrix} 120s^{5} + 274s^{4} + 225s^{3} + 85s^{2} + 15s + 1 \\ \end{bmatrix}}$$

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}}{\begin{bmatrix} 120\left( \text{jω} \right)^{5} + 274\left( \text{jω} \right)^{4} + 225\left( \text{jω} \right)^{3} + 85\left( \text{jω} \right)^{2} + 15\left( \text{jω} \right) + 1 \\ \end{bmatrix}}$$

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}}{\begin{bmatrix} 120j\omega^{5} + 274\omega^{4} - 225\text{jω}^{3} - 85\omega^{2} + 15j\omega + 1 \\ \end{bmatrix}}$$

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right) - j\left( 120\omega^{5} - 225\omega^{3} + 15\omega \right) \\ \end{bmatrix}}{\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right) + j\left( 120\omega^{5} - 225\omega^{3} + 15\omega \right) \\ \end{bmatrix}\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right) - j\left( 120\omega^{5} - 225\omega^{3} + 15\omega \right) \\ \end{bmatrix}}$$

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right) - j\left( 120\omega^{5} - 225\omega^{3} + 15\omega \right) \\ \end{bmatrix}}{\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2} \\ \end{bmatrix}}$$

$$P_{\text{uo}}\left( \omega \right) = \frac{k_{p}\left( 274\omega^{4} - 85\omega^{2} + 1 \right)}{\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2} \\ \end{bmatrix}}$$

$$Q_{\text{uo}}\left( \omega \right) = \frac{- k_{p}\left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)}{\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2} \\ \end{bmatrix}}$$

Warunek granicy stabilności

$$\left\{ \begin{matrix} P_{\text{uo}}\left( \omega \right) = - 1 \\ Q_{\text{uo}}\left( \omega \right) = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \frac{k_{p}\left( 274\omega^{4} - 85\omega^{2} + 1 \right)}{\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2} \\ \end{bmatrix}} = - 1 \\ \frac{- k_{p}\left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)}{\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2} \\ \end{bmatrix}} = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \frac{k_{p}\left( 274\omega^{4} - 85\omega^{2} + 1 \right)}{\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2} \\ \end{bmatrix}} = - 1 \\ - k_{p}\left( 120\omega^{5} - 225\omega^{3} + 15\omega \right) = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \frac{k_{p}\left( 274\omega^{4} - 85\omega^{2} + 1 \right)}{\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2} \\ \end{bmatrix}} = - 1 \\ 120\omega^{5} - 225\omega^{3} + 15\omega = 0 \\ \end{matrix} \right.\ $$

15ω(8ω⁴−15ω²+1) = 0

Δ = 225 − 4 • 8 = 193

ω₁² = 0, 069222       i          ω₂² = 1, 805778

ω₁ = 0, 2631       i          ω₂ = 1, 3438

$$P_{\text{uo}}\left( 0,2631 \right) = \frac{k_{p}\left( 274\omega^{4} - 85\omega^{2} + 1 \right)}{\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2} \\ \end{bmatrix}} = - 1$$

$$\frac{k_{p}\left( 274\left( 0,2631 \right)^{4} - 85\left( 0,2631 \right)^{2} + 1 \right)}{\begin{bmatrix} \left( 274\left( 0,2631 \right)^{4} - 85\left( 0,2631 \right)^{2} + 1 \right)^{2} + \left( 120\left( 0,2631 \right)^{5} - 225\left( 0,2631 \right)^{3} + 15\left( 0,2631 \right) \right)^{2} \\ \end{bmatrix}} = - 1$$

$$\frac{k_{p}}{\left( 274\left( 0,2631 \right)^{4} - 85\left( 0,2631 \right)^{2} + 1 \right)} = - 1$$

k_p = −274(0,2631)⁴ + 85(0,2631)² − 1

k_p = 3, 571

$$\left\{ \begin{matrix} \omega = 0,2631 \\ \left( k_{p} \right)_{\text{kr}} = 3,571 \\ \end{matrix} \right.\ $$

Warunek: zapas modułu ΔL=6dB

ΔL = 20lgΔM = −6

$$\Delta M = 10^{- \frac{6}{20}}$$

ΔM = 0, 5

$$\left\{ \begin{matrix} P_{\text{uo}}\left( \omega \right) = - 0,5 \\ Q_{\text{uo}}\left( \omega \right) = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \frac{k_{p}\left( 274\omega^{4} - 85\omega^{2} + 1 \right)}{\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2} \\ \end{bmatrix}} = - 0,5 \\ \frac{- k_{p}\left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)}{\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2} \\ \end{bmatrix}} = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \omega = 0,2631 \\ k_{p} = 1,7855 \\ \end{matrix} \right.\ $$

Warunek: zapas fazy Δϕ=60 ̊

$$\left\{ \begin{matrix} M\left( \omega \right) = 1 \\ \tan\varphi = \frac{Q\left( \omega \right)}{P\left( \omega \right)} = \sqrt{3} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \sqrt{{P\left( \omega \right)}^{2} + {Q\left( \omega \right)}^{2}} = 1 \\ \tan\varphi = \frac{Q\left( \omega \right)}{P\left( \omega \right)} = \sqrt{3} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \sqrt{\begin{matrix} \left( \frac{k_{p}\left( 274\omega^{4} - 85\omega^{2} + 1 \right)}{\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2} \\ \end{bmatrix}} \right)^{2} + \\ \left( \frac{- k_{p}\left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)}{\begin{bmatrix} \left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2} \\ \end{bmatrix}} \right)^{2} \\ \end{matrix}} = 1 \\ \tan\varphi = \frac{Q\left( \omega \right)}{P\left( \omega \right)} = \sqrt{3} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \frac{k_{p}}{\sqrt{\left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2}}} = 1 \\ \frac{- \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)}{274\omega^{4} - 85\omega^{2} + 1} = \sqrt{3} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = \sqrt{\left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2}} \\ 120\omega^{5} - 225\omega^{3} + 15\omega = - \sqrt{3}\left( 274\omega^{4} - 85\omega^{2} + 1 \right) \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = \sqrt{\left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2}} \\ 120\omega^{5} + 274\sqrt{3}\omega^{4} - 225\omega^{3} - 85\sqrt{3}\omega^{2} + 15\omega + \sqrt{3} = 0 \\ \end{matrix} \right.\ $$

Polecenie w Matlab

>> roots([120 274*sqrt(3) -225 -85*sqrt(3) 15 sqrt(3)])

ans =

-4.3215

0.7104

-0.4263

ω= 0.1542

-0.0715

$$\left\{ \begin{matrix} k_{p} = \sqrt{\left( 274\omega^{4} - 85\omega^{2} + 1 \right)^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)^{2}} \\ \omega = \ 0.1542 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = 1.731 \\ \omega = \ 0.1542 \\ \end{matrix} \right.\ $$

Przykład:

Obiekt inercyjny piątego rzędu i regulator PI

$$G\left( s \right) = \frac{k}{\left( T_{1}s + 1 \right)\left( T_{2}s + 1 \right)\left( T_{3}s + 1 \right)\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

$$R\left( s \right) = k_{p}\left( 1 + \frac{1}{T_{i}s} \right)$$

$$G_{\text{uo}}\left( s \right) = \frac{kk_{p}\left( 1 + \frac{1}{T_{i}s} \right)}{\left( T_{1}s + 1 \right)\left( T_{2}s + 1 \right)\left( T_{3}s + 1 \right)\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

Dane:

k = 1,   T₁ = 1,   T₂ = 2,   T₃ = 3,   T₄ = 4,   T₅ = 5

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\left( 1 + \frac{1}{T_{i}\text{jω}} \right)}{\begin{bmatrix} 120j\omega^{5} + 274\omega^{4} - 225\text{jω}^{3} - 85\omega^{2} + 15j\omega + 1 \\ \end{bmatrix}}$$

a = 274ω⁴ − 85ω² + 1

b = 120ω⁵ − 225ω³ + 15ω

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\left( a - jb \right)}{\left( a^{2} + b^{2} \right)} \bullet \left( 1 - j\frac{1}{T_{i}\omega} \right)$$

$$G_{\text{uo}}\left( \text{jω} \right) = k_{p}\frac{a - \frac{b}{T_{i}\omega} - j\frac{a}{T_{i}\omega} - jb}{\left( a^{2} + b^{2} \right)}$$

$$G_{\text{uo}}\left( \text{jω} \right) = k_{p}\frac{a - \frac{b}{T_{i}\omega} - j\left( \frac{a}{T_{i}\omega} + b \right)}{\left( a^{2} + b^{2} \right)}$$

$$P_{\text{uo}}\left( \omega \right) = k_{p}\frac{a - \frac{b}{T_{i}\omega}}{\left( a^{2} + b^{2} \right)}$$

$$Q_{\text{uo}}\left( \omega \right) = - k_{p}\frac{\frac{a}{T_{i}\omega} + b}{\left( a^{2} + b^{2} \right)}$$

Warunek granicy stabilności

$$\left\{ \begin{matrix} P_{\text{uo}}\left( \omega \right) = - 1 \\ Q_{\text{uo}}\left( \omega \right) = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}\frac{a - \frac{b}{T_{i}\omega}}{\left( a^{2} + b^{2} \right)} = - 1 \\ - k_{p}\frac{\frac{a}{T_{i}\omega} + b}{\left( a^{2} + b^{2} \right)} = 0 \\ \end{matrix} \right.\ $$

$$T_{i}\omega = - \frac{a}{b}$$

$$k_{p}\frac{a + \frac{b^{2}}{a}}{\left( a^{2} + b^{2} \right)} = - 1$$

$$k_{p}\frac{1}{a} = - 1$$

k_p = −a

$$\left\{ \begin{matrix} k_{p} = - a \\ T_{i}\omega = - \frac{a}{b} \\ \end{matrix} \right.\ $$

$$\frac{k_{p}}{T_{i}\omega} = b$$

$$\frac{k_{p}}{T_{i}\omega} = \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)$$

Szukamy ekstremum funkcji (duże wzmocnienie i duże całkowanie-mała stała czasowa to stosunek ma być największy)

$$\frac{k_{p}}{T_{i}} = 15\left( 8\omega^{6} - 15\omega^{4} + \omega^{2} \right)$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = 15\left( 48\omega^{5} - 60\omega^{3} + 2\omega \right)$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = 30\omega\left( 24\omega^{4} - 30\omega^{2} + 1 \right)$$

Δ = 900 − 4 • 24 = 804

ω₁² = 0, 034273       i          ω₂² = 1, 215727

ω₁ = 0, 18513       i          ω₂ = 1, 1026

$$\left\{ \begin{matrix} k_{p} = - a \\ T_{i}\omega = - \frac{a}{b} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = - \left( 274\omega^{4} - 85\omega^{2} + 1 \right) \\ T_{i}\omega = - \frac{274\omega^{4} - 85\omega^{2} + 1}{120\omega^{5} - 225\omega^{3} + 15\omega} \\ \end{matrix} \right.\ $$

Dla ω=0,18513

$$\left\{ \begin{matrix} \left( k_{p} \right)_{\text{kr}} = 1,59 \\ T_{i} = 6,25 \\ \end{matrix} \right.\ $$

Warunek: zapas modułu ΔL=ΔdB

ΔL = 20lgΔM = −Δ

$$\Delta M = 10^{- \frac{\Delta}{20}}$$

$$\left\{ \begin{matrix} P_{\text{uo}}\left( \omega \right) = - 1 + \Delta M \\ Q_{\text{uo}}\left( \omega \right) = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}\frac{a - \frac{b}{T_{i}\omega}}{\left( a^{2} + b^{2} \right)} = \Delta M - 1 \\ {- k}_{p}\frac{\frac{a}{T_{i}\omega} + b}{\left( a^{2} + b^{2} \right)} = 0 \\ \end{matrix} \right.\ $$

$$T_{i}\omega = - \frac{a}{b}$$

$$k_{p}\frac{a + \frac{b^{2}}{a}}{\left( a^{2} + b^{2} \right)} = \Delta M - 1$$

$$k_{p}\frac{1}{a} = \Delta M - 1$$

k_p = −a(1−ΔM)

k_p = −(1−ΔM)(274ω⁴−85ω²+1)

$$\left\{ \begin{matrix} k_{p} = - a\left( 1 - \Delta M \right) \\ T_{i}\omega = - \frac{a}{b} \\ \end{matrix} \right.\ $$

$$\frac{k_{p}}{T_{i}\omega} = b\left( 1 - \Delta M \right)$$

$$\frac{k_{p}}{T_{i}\omega} = \left( 1 - \Delta M \right)\left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)$$

Szukamy ekstremum funkcji (duże wzmocnienie i duże całkowanie-mała stała czasowa to stosunek ma być największy)

$$\frac{k_{p}}{T_{i}} = 15\left( 1 - \Delta M \right)\left( 8\omega^{6} - 15\omega^{4} + \omega^{2} \right)$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = 15\left( 1 - \Delta M \right)\left( 48\omega^{5} - 60\omega^{3} + 2\omega \right)$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = 30\omega\left( 1 - \Delta M \right)\left( 24\omega^{4} - 30\omega^{2} + 1 \right)$$

Δ = 900 − 4 • 24 = 804

ω₁² = 0, 034273       i          ω₂² = 1, 215727

ω₁ = 0, 18513       i          ω₂ = 1, 1026

$$\left\{ \begin{matrix} k_{p} = - a\left( 1 - \Delta M \right) \\ T_{i}\omega = - \frac{a}{b} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = - \left( 1 - \Delta M \right)\left( 274\omega^{4} - 85\omega^{2} + 1 \right) \\ T_{i}\omega = - \frac{274\omega^{4} - 85\omega^{2} + 1}{120\omega^{5} - 225\omega^{3} + 15\omega} \\ \end{matrix} \right.\ $$

Dla ΔL=6dB i ω=0,18513

$$\left\{ \begin{matrix} k_{p} = 0,795679 \\ T_{i} = 6,25 \\ \end{matrix} \right.\ $$

Warunek: zapas fazy Δϕ

$$\left\{ \begin{matrix} M\left( \omega \right) = 1 \\ \tan\text{Δφ} = \frac{Q\left( \omega \right)}{P\left( \omega \right)} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \sqrt{{P\left( \omega \right)}^{2} + {Q\left( \omega \right)}^{2}} = 1 \\ \tan\text{Δφ} = \frac{Q\left( \omega \right)}{P\left( \omega \right)} \\ \end{matrix} \right.\ $$

a = 274ω⁴ − 85ω² + 1

b = 120ω⁵ − 225ω³ + 15ω

$$\left\{ \begin{matrix} k_{p}\sqrt{\frac{1 + \frac{1}{\left( T_{i}\omega \right)^{2}}}{a^{2} + b^{2}}} = 1 \\ \tan\text{Δφ} = - \frac{\frac{a}{T_{i}\omega} + b}{a - \frac{b}{T_{i}\omega}} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2}\left( 1 + \frac{1}{\left( T_{i}\omega \right)^{2}} \right) = a^{2} + b^{2} \\ \operatorname{a\bullet tan}\text{Δφ} - \frac{b}{T_{i}\omega}\tan\text{Δφ} = - \frac{a}{T_{i}\omega} - b \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2}\left( 1 + \frac{1}{\left( T_{i}\omega \right)^{2}} \right) = a^{2} + b^{2} \\ \operatorname{a\bullet tan}\text{Δφ} + b = \frac{1}{T_{i}\omega}\left( b\tan\text{Δφ} - a \right) \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2} = \frac{a^{2} + b^{2}}{1 + \frac{1}{\left( T_{i}\omega \right)^{2}}} \\ T_{i}\omega = \frac{b\tan\text{Δφ} - a}{\operatorname{a\bullet tan}\text{Δφ} + b} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2} = \frac{a^{2} + b^{2}}{1 + \frac{1}{\left( \frac{b\tan\text{Δφ} - a}{\operatorname{a\bullet tan}\text{Δφ} + b} \right)^{2}}} \\ T_{i}\omega = \frac{b\tan\text{Δφ} - a}{\operatorname{a\bullet tan}\text{Δφ} + b} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2} = \frac{a^{2} + b^{2}}{\frac{\left( b\tan\text{Δφ} - a \right)^{2} + \left( \operatorname{a\bullet tan}\text{Δφ} + b \right)^{2}}{\left( b\tan\text{Δφ} - a \right)^{2}}} \\ T_{i}\omega = \frac{b\tan\text{Δφ} - a}{\operatorname{a\bullet tan}\text{Δφ} + b} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2} = \frac{\left( a^{2} + b^{2} \right)\left( b\tan\text{Δφ} - a \right)^{2}}{\left( b\tan\text{Δφ} - a \right)^{2} + \left( \operatorname{a\bullet tan}\text{Δφ} + b \right)^{2}} \\ T_{i}\omega = \frac{b\tan\text{Δφ} - a}{\operatorname{a\bullet tan}\text{Δφ} + b} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2} = \frac{\left( a^{2} + b^{2} \right)\left( b\tan\text{Δφ} - a \right)^{2}}{\left( a^{2} + b^{2} \right)\left\lbrack 1 + \left( \tan\text{Δφ} \right)^{2} \right\rbrack} \\ T_{i}\omega = \frac{b\tan\text{Δφ} - a}{\operatorname{a\bullet tan}\text{Δφ} + b} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = \frac{b\tan\text{Δφ} - a}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}} \\ T_{i}\omega = \frac{b\tan\text{Δφ} - a}{\operatorname{a\bullet tan}\text{Δφ} + b} \\ \end{matrix} \right.\ $$

$$\frac{k_{p}}{T_{i}\omega} = \frac{\operatorname{a\bullet tan}\text{Δφ} + b}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}$$

$$\frac{k_{p}}{T_{i}} = \frac{\operatorname{a\omega\bullet tan}\text{Δφ} + b\omega}{\sqrt{1 + \left( \tan{\Delta\varphi} \right)^{2}}}$$

$$\frac{k_{p}}{T_{i}} = \frac{\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}a\omega + \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\text{bω}$$

$$\frac{k_{p}}{T_{i}} = \frac{\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 274\omega^{5} - 85\omega^{3} + \omega \right) + \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 120\omega^{6} - 225\omega^{4} + 15\omega^{2} \right)$$

Liczymy ekstremum

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = \frac{\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 1370\omega^{4} - 255\omega^{2} + 1 \right) + \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 720\omega^{5} - 900\omega^{3} + 30\omega \right)$$

Dla Δϕ=60 ̊

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = \frac{\sqrt{3}}{2}\left( 1370\omega^{4} - 255\omega^{2} + 1 \right) + 15\left( 24\omega^{5} - 30\omega^{3} + \omega \right)$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = 360\omega^{5} + 685{\sqrt{3}\omega}^{4} - 450\omega^{3} - 127,5\sqrt{3}\omega^{2} + 15\omega + \frac{\sqrt{3}}{2}$$

>> roots([360 685*sqrt(3) -450 -127.5*sqrt(3) 15 sqrt(3)/2])

ans =

-3.5951

0.5668

-0.3246

0.0953

-0.0381

$$\left\{ \begin{matrix} k_{p} = \frac{b\tan\text{Δφ} - a}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}} \\ T_{i}\omega = \frac{b\tan\text{Δφ} - a}{\operatorname{a\bullet tan}\text{Δφ} + b} \\ \end{matrix} \right.\ $$

Dla ω=0,0953

$$\left\{ \begin{matrix} k_{p} = 1,172 \\ T_{i} = 14,734 \\ \end{matrix} \right.\ $$

Badania z regulatorem PID

Przykład:

Obiekt inercyjny piątego rzędu aproksymowany transmitancją zastępczą i regulator P

$$G\left( s \right) = \frac{k}{\left( T_{1}s + 1 \right)\left( T_{2}s + 1 \right)\left( T_{3}s + 1 \right)\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

R(s) = k_p

Dane:

k = 1,   T₁ = 1,   T₂ = 2,   T₃ = 3,   T₄ = 4,   T₅ = 5

$$G\left( s \right) = \frac{k}{\left( s + 1 \right)\left( 2s + 1 \right)\left( 3s + 1 \right)\left( 4s + 1 \right)\left( 5s + 1 \right)}$$

$$G\left( s \right) = \frac{1 \bullet e^{- \tau s}}{T_{z}s + 1}$$

$$G_{\text{uo}}\left( s \right) = \frac{k_{p}e^{- \tau s}}{T_{z}s + 1}$$

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}e^{- \tau j\omega}}{T_{z}j\omega + 1}$$

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\left( \cos{\tau\omega - j\sin\text{τω}} \right)}{1 + T_{z}\text{jω}} \bullet \frac{1 - T_{z}\text{jω}}{1 - T_{z}\text{jω}}$$

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\left( \cos{\tau\omega - T_{z}\omega\sin\text{τω} - j\sin\text{τω}} - T_{z}\text{jω}\cos\text{τω} \right)}{1 + T_{z}^{2}\omega^{2}}$$

$$P_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\left( \cos\text{τω} - T_{z}\omega\sin\text{τω} \right)}{1 + T_{z}^{2}\omega^{2}}$$

$$Q_{\text{uo}}\left( \text{jω} \right) = - \frac{k_{p}\left( \sin\text{τω} + T_{z}\omega\cos\text{τω} \right)}{1 + T_{z}^{2}\omega^{2}}$$

$$M\left( \omega \right) = \frac{k_{p}}{\sqrt{1 + T_{z}^{2}\omega^{2}}}$$

$$tg\varphi = - \frac{T_{z}\omega\cos\text{τω} + \sin\text{τω}}{\cos\text{τω} - \operatorname{\omega sin}\text{τω}}$$

Warunek granicy stabilności

$$\left\{ \begin{matrix} P_{\text{uo}}\left( \omega \right) = - 1 \\ Q_{\text{uo}}\left( \omega \right) = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \frac{k_{p}\left( \cos\text{τω} - T_{z}\omega\sin\text{τω} \right)}{1 + T_{z}^{2}\omega^{2}} = - 1 \\ - \frac{k_{p}\left( \sin\text{τω} + T_{z}\omega\cos\text{τω} \right)}{1 + T_{z}^{2}\omega^{2}} = 0 \\ \end{matrix} \right.\ $$

sinτω + T_zωcosτω = 0

T_zωcosτω = -sinτω

$$T_{z}\omega = \frac{\sin\text{τω}}{\cos\text{τω}}$$

T_zω = tanτω

tanτω − T_zω = 0

tan6ω − 15ω = 0

Równanie uwikłane, z rozwiązania cyfrowego ω=0,210733

$$\left\{ \begin{matrix} \frac{k_{p}\left( \cos\text{τω} - T_{z}\omega\sin\text{τω} \right)}{1 + T_{z}^{2}\omega^{2}} = - 1 \\ \omega = 0,210733 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = - \frac{1 + T_{z}^{2}\omega^{2}}{\cos\text{τω} - T_{z}\omega\sin\text{τω}} \\ \omega = 0,210733 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \left( k_{p} \right)_{\text{kr}} = 4,0528 \\ \omega = 0,210733 \\ \end{matrix} \right.\ $$

Warunek: zapas modułu ΔL=6dB

ΔL = 20lgΔM = −6

$$\Delta M = 10^{- \frac{6}{20}}$$

ΔM = 0, 5

$$\left\{ \begin{matrix} P_{\text{uo}}\left( \omega \right) = - 1 + \Delta M \\ Q_{\text{uo}}\left( \omega \right) = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \frac{k_{p}\left( \cos\text{τω} - T_{z}\omega\sin\text{τω} \right)}{1 + T_{z}^{2}\omega^{2}} = - 1 + \Delta M \\ - \frac{k_{p}\left( \sin\text{τω} + T_{z}\omega\cos\text{τω} \right)}{1 + T_{z}^{2}\omega^{2}} = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = 2,0264 \\ \omega = 0,210733 \\ \end{matrix} \right.\ $$

Warunek: zapas fazy Δϕ=60 ̊

$$\left\{ \begin{matrix} M\left( \omega \right) = 1 \\ \tan\text{Δφ} = \frac{Q\left( \omega \right)}{P\left( \omega \right)} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \frac{k_{p}}{\sqrt{1 + T_{z}^{2}\omega^{2}}} = 1 \\ \tan\text{Δφ} = - \frac{T_{z}\omega\cos\text{τω} + \sin\text{τω}}{\cos\text{τω} - \operatorname{\omega sin}\text{τω}} \\ \end{matrix} \right.\ $$

tanΔφcosτω − ωtanΔφsinτω = −T_zωcosτω − sinτω

tanΔφcosτω + T_zωcosτω = ωtanΔφsinτω − sinτω

(tanΔφ+T_zω)cosτω = sinτω

$$\frac{\tan\text{Δφ} + T_{z}\omega}{T_{z}\omega\tan\text{Δφ} - 1} = \frac{\sin\text{τω}}{\cos\text{τω}}$$

$$\frac{\tan\text{Δφ} + T_{z}\omega}{T_{z}\omega\tan\text{Δφ} - 1} = \tan\text{τω}$$

Z rozwiązania cyfrowego ω=0,155

$$\left\{ \begin{matrix} k_{p} = \sqrt{1 + T_{z}^{2}\omega^{2}} \\ \omega = 0,155 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = 2,530934 \\ \omega = 0,155 \\ \end{matrix} \right.\ $$

Stabilność – kryterium Hurwitza

Badanie stabilności układu regulacji na podstawie transmitancji operatorowej.

$$G\left( s \right) = \frac{k}{a_{n}s^{n} + a_{n - 1}s^{n - 1} + ...\ldots\ldots + a_{3}s^{3} + a_{2}s^{2} + a_{1}s + a_{0}}$$

Macierz Hurwitza

$$\Delta_{n} = \left| \begin{matrix} \begin{matrix} \begin{matrix} a_{n - 1} & a_{n} & 0 \\ a_{n - 3} & a_{n - 2} & a_{n - 1} \\ a_{n - 5} & a_{n - 4} & a_{n - 3} \\ \end{matrix} \\ \begin{matrix} a_{n - 7} & a_{n - 6} & a_{n - 5} \\ a_{n - 9} & a_{n - 8} & a_{n - 7} \\ \ldots & \ldots & \ldots \\ \end{matrix} \\ \end{matrix} & \begin{matrix} 0 \\ a_{n} \\ \begin{matrix} a_{n - 2} \\ a_{n - 4} \\ \begin{matrix} a_{n - 6} \\ \ldots \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 & \ldots \\ \end{matrix} \\ \begin{matrix} 0 & \ldots \\ \end{matrix} \\ \begin{matrix} \begin{matrix} a_{n - 1} & \ldots \\ \end{matrix} \\ \begin{matrix} a_{n - 3} & \ldots \\ \end{matrix} \\ \begin{matrix} \begin{matrix} a_{n - 5} & \ldots \\ \end{matrix} \\ \begin{matrix} \ldots & \ldots \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right|$$

1. Warunek konieczny - wszystkie wyrazy równania charakterystycznego istnieją i są dodatnie

a_n > 0,    a_n − 1 > 0,   ...……   a₃ > 0,    a₂ > 0      a₁ > 0,      a₀ > 0

1. Warunek dostateczny – wszystkie podwyznaczniki wyznacznika głównego macierzy Hurwitza są większe od zera

Δ_n > 0,    Δ_n − 1 > 0,   ...……   Δ₃ > 0,    Δ₂ > 0      Δ₁ > 0

Jeżeli któryś ze współczynników równania charakterystycznego jest ujemny lub równy zero albo któryś z podwyznaczników jest ujemny lub równy zero to układ jest niestabilny.

Ponieważ Δ_n = a₀Δ_n − 1 i Δ₁ = a_n − 1, to niecelowe jest sprawdzanie dodatności tych wyznaczników. Konieczne jest sprawdzanie dodatności wyznaczników od Δ₂ do Δ_n-1.

Jeżeli któryś z podwyznaczników jest równy zero to układ jest na granicy stabilności – oscylacje o stałej amplitudzie.

Kryterium Hurwitza nie podaje żadnych progów stabilności.

Przykład:

$$G\left( s \right) = \frac{k}{\left( T_{1}s + 1 \right)\left( T_{2}s + 1 \right)\left( T_{3}s + 1 \right)\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

Dane:

k = 1,   T₁ = 1,   T₂ = 2,   T₃ = 3,   T₄ = 4,   T₅ = 5

$$G\left( s \right) = \frac{1}{120s^{5} + 274s^{4} + 225s^{3} + 85s^{2} + 15s + 1}$$

a₅ = 120,    a₄ = 274,     a₃ = 225,    a₂ = 85      a₁ = 15,      a₀ = 1

$$\Delta_{5} = \left| \begin{matrix} \begin{matrix} \begin{matrix} a_{4} & a_{5} & 0 \\ a_{2} & a_{3} & a_{4} \\ a_{0} & a_{1} & a_{2} \\ \end{matrix} \\ \begin{matrix} 0 & 0 & a_{0} \\ 0 & 0 & 0 \\ \end{matrix} \\ \end{matrix} & \begin{matrix} 0 \\ a_{5} \\ \begin{matrix} a_{3} \\ a_{1} \\ \begin{matrix} 0 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} \begin{matrix} a_{4} \\ \end{matrix} \\ \begin{matrix} a_{2} \\ \end{matrix} \\ \begin{matrix} \begin{matrix} a_{0} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right|$$

$$\Delta_{5} = \left| \begin{matrix} \begin{matrix} \begin{matrix} 274 & 120 & 0 \\ 85 & 225 & 274 \\ \begin{matrix} 1 \\ 0 \\ 0 \\ \end{matrix} & \begin{matrix} 15 \\ 0 \\ 0 \\ \end{matrix} & \begin{matrix} 85 \\ 1 \\ 0 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} 0 \\ 120 \\ \begin{matrix} 225 \\ 15 \\ \begin{matrix} 0 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 274 \\ \end{matrix} \\ \begin{matrix} 85 \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 1 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right|$$

$$\Delta_{2} = \left| \begin{matrix} 274 & 120 \\ 85 & 225 \\ \end{matrix} \right| = 274 \bullet 225 - 120 \bullet 85 = 51450$$

$$\Delta_{3} = \left| \begin{matrix} \begin{matrix} \begin{matrix} 274 & 120 & 0 \\ 85 & 225 & 274 \\ \begin{matrix} 1 \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} & \begin{matrix} 85 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| = 274 \bullet 225 \bullet 85 + 1 \bullet 120 \bullet 274 - 274 \bullet 15 \bullet 274 - 85 \bullet 120 \bullet 85 = 3279990$$

$$\Delta_{4} = \left| \begin{matrix} \begin{matrix} \begin{matrix} 274 & 120 & 0 \\ 85 & 225 & 274 \\ \begin{matrix} 1 \\ 0 \\ \end{matrix} & \begin{matrix} 15 \\ 0 \\ \end{matrix} & \begin{matrix} 85 \\ 1 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} 0 \\ 120 \\ \begin{matrix} 225 \\ 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| = \left( - 1 \right)^{1 + 1} \bullet 274 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 225 & 274 & 120 \\ 15 & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| + \left( - 1 \right)^{1 + 2} \bullet 120 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 85 & 274 & 120 \\ 1 & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| =$$

$$\Delta_{4} = 274 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 225 & 274 & 120 \\ 15 & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| - 120 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 85 & 274 & 120 \\ 1 & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| =$$

Δ₄ = 274(225•85•15+15•120−225•225−15•274•15) − 120(85•85•15+120−85•225−274•15) = 38102400

Układ otwarty jest stabilny, wszystkie warunki są spełnione

Badanie granicy stabilności układu zamkniętego:

$$G\left( s \right) = \frac{1}{120s^{5} + 274s^{4} + 225s^{3} + 85s^{2} + 15s + 1}$$

R(s) = k_p

$$G_{z}\left( s \right) = \frac{k_{p}\frac{1}{120s^{5} + 274s^{4} + 225s^{3} + 85s^{2} + 15s + 1}}{1 + k_{p}\frac{1}{120s^{5} + 274s^{4} + 225s^{3} + 85s^{2} + 15s + 1}}$$

$$G_{z}\left( s \right) = \frac{k_{p}}{120s^{5} + 274s^{4} + 225s^{3} + 85s^{2} + 15s + 1 + k_{p}}$$

a₅ = 120,    a₄ = 274,     a₃ = 225,    a₂ = 85      a₁ = 15,      a₀ = 1 + k_p

$$\Delta_{5} = \left| \begin{matrix} \begin{matrix} \begin{matrix} a_{4} & a_{5} & 0 \\ a_{2} & a_{3} & a_{4} \\ a_{0} & a_{1} & a_{2} \\ \end{matrix} \\ \begin{matrix} 0 & 0 & a_{0} \\ 0 & 0 & 0 \\ \end{matrix} \\ \end{matrix} & \begin{matrix} 0 \\ a_{5} \\ \begin{matrix} a_{3} \\ a_{1} \\ \begin{matrix} 0 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} \begin{matrix} a_{4} \\ \end{matrix} \\ \begin{matrix} a_{2} \\ \end{matrix} \\ \begin{matrix} \begin{matrix} a_{0} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right|$$

$$\Delta_{5} = \left| \begin{matrix} \begin{matrix} \begin{matrix} 274 & 120 & 0 \\ 85 & 225 & 274 \\ \begin{matrix} 1 + k_{p} \\ 0 \\ 0 \\ \end{matrix} & \begin{matrix} 15 \\ 0 \\ 0 \\ \end{matrix} & \begin{matrix} 85 \\ 1 + k_{p} \\ 0 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} 0 \\ 120 \\ \begin{matrix} 225 \\ 15 \\ \begin{matrix} 0 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 274 \\ \end{matrix} \\ \begin{matrix} 85 \\ \end{matrix} \\ 1 + k_{p} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right|$$

$$\Delta_{2} = \left| \begin{matrix} 274 & 120 \\ 85 & 225 \\ \end{matrix} \right| = 274 \bullet 225 - 120 \bullet 85 = 51450$$

$$\Delta_{3} = \left| \begin{matrix} \begin{matrix} \begin{matrix} 274 & 120 & 0 \\ 85 & 225 & 274 \\ \begin{matrix} 1 + k_{p} \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} & \begin{matrix} 85 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| = 274 \bullet 225 \bullet 85 + \left( 1 + k_{p} \right) \bullet 120 \bullet 274 - 274 \bullet 15 \bullet 274 - 85 \bullet 120 \bullet 85 = 5240250 + \left( 1 + k_{p} \right)32880 - 1126140 - 867000 = 3247110 + \left( 1 + k_{p} \right)32880$$

k_p ≥ −97, 756

$$\Delta_{4} = \left| \begin{matrix} \begin{matrix} \begin{matrix} 274 & 120 & 0 \\ 85 & 225 & 274 \\ \begin{matrix} 1 + k_{p} \\ 0 \\ \end{matrix} & \begin{matrix} 15 \\ 0 \\ \end{matrix} & \begin{matrix} 85 \\ 1 + k_{p} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} 0 \\ 120 \\ \begin{matrix} 225 \\ 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| = \left( - 1 \right)^{1 + 1} \bullet 274 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 225 & 274 & 120 \\ 15 & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 + k_{p} \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| + \left( - 1 \right)^{1 + 2} \bullet 120 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 85 & 274 & 120 \\ 1 + k_{p} & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 + k_{p} \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| =$$

$$\Delta_{4} = 274 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 225 & 274 & 120 \\ 15 & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 + k_{p} \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| - 120 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 85 & 274 & 120 \\ 1 + k_{p} & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 + k_{p} \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| =$$

Δ₄ = 274(225•85•15+15•120(1+k_p)−225•225(1+k_p)−15•274•15) − 120(85•85•15+120(1+k_p)(1+k_p)−85•225(1+k_p)−274•15(1+k_p)) = 274(286875+1800(1+k_p)−50625(1+k_p)−61650) − 120(108375+120(1+k_p)(1+k_p)−19125(1+k_p)−4110(1+k_p)) = 274(225225−48825(1+k_p)) − 120(108375+120(1+k_p)(1+k_p)−23235(1+k_p)) = 48706650 − 10589850(1+k_p) − 14400(1+k_p)(1+k_p)

1 + k_p = 4, 571     i       1 + k_p = −734

Wzmocnienie krytyczne wynosi (k_p)_kr = 3, 571